Combining root laws

It is also important to remember something fundamental, the important rules and the laws of roots.

A root is the inverse operation of exponentiation,
it is denoted by the symbol $√$ and is equivalent to the power of $0.5$.
If a small number appears on the left side, it indicates the order of the root.

1. The result of the root will always be positive! A negative result will never be obtained. We can get a result of 0.
2. A root is essentially a power of one-half. We can say that: $\sqrt a=a^{ 1 \over 2}$
3. Root comes before the four arithmetic operations. First, perform the root and only then arrange the arithmetic operations.

When the root appears over the entire product, we can break down each factor and apply the root while leaving the multiplication sign between the factors.
We will formulate it as a formula:
$\sqrt{(a\cdot b)}=\sqrt{a}\cdot\sqrt{b}$

When the root appears on the entire quotient (on the entire fraction), we can break down each factor and apply the root to it while leaving the division sign (the fraction line) between the factors.
We can formulate it as a formula:
$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

The root on an additional root, we multiply the order of the first root by the order of the second root, and the order we obtained is applied as a root to our number. (Similar to the rule of power on power)
We will formulate this as a rule:
$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\cdot m]{a}$

And now we will practice exercises that combine the laws of exponents:
$\sqrt{9\cdot16}+\sqrt{\frac{25}{36}}\cdot\sqrt{6-2}=$

Solution:
According to what we learned, we must first handle the roots according to the order of operations, so we will start from the beginning of the exercise to handle each root.
Let's start with the root $\sqrt{9\cdot16}$
The root here is a root of a product, and according to the formula for a root with a product, we can separate the two numbers to solve more easily
$\sqrt{(a\cdot b)}=\sqrt{a}\cdot\sqrt{b}$
In our exercise, this means
$\sqrt{9\cdot16}=\sqrt{9}\cdot\sqrt{16}$
We will rewrite the exercise and continue
$\sqrt{9}\cdot\sqrt{16}+\sqrt{\frac{25}{36}}\cdot\sqrt{6-2}=$
Great! Now let's move on to the second root that needs handling
$\sqrt{\frac{25}{36}}$
The root here is a root of a quotient, and according to the formula, we can separate the numerator and the denominator to make it easier to solve.
The formula is:
$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

And in our exercise:
$\sqrt{\frac{25}{36}}=\frac{\sqrt{25}}{\sqrt{36}}$
We will rewrite the exercise and continue:
$\sqrt{9}\cdot\sqrt{16}+\frac{\sqrt{25}}{\sqrt{36}}\cdot\sqrt{6-2}=$
Now let's move to the root
$\sqrt{6-2}$
Here, no formula is needed, and we can simply solve what is inside the root to get:
$\sqrt{6-2}=\sqrt4$
Now we will rewrite the exercise:
$\sqrt{9}\cdot\sqrt{16}+\frac{\sqrt{25}}{\sqrt{36}}\cdot\sqrt{4}=$
Note, according to what we learned, the root comes before arithmetic operations, so we will first solve the simple roots we created for ourselves:
$\sqrt9=3$
$\sqrt{16}=4$

$\frac{\sqrt{25}}{\sqrt{36}}=\frac{5}{6}$
$\sqrt4=2$
Now let's rewrite the exercise after we have eliminated and solved all the roots:
$3\cdot4+\frac{5}{6}\cdot2=$
We will continue to solve according to the order of operations.
Multiplication comes before addition, therefore:
$12+\frac{10}{6}=13\frac{4}{6}=13\frac{2}{3}$